Summation of Series by the Method of Induction | 数学归纳法求和级数

📚 Summation of Series by the Method of Induction | 数学归纳法求和级数

A series is the sum of the terms of a sequence. When a formula for such a sum is known or conjectured, mathematical induction provides a rigorous way to prove that it is true for every positive integer (n). This technique is a core part of AQA A-Level Mathematics because it combines structure with algebraic fluency.

级数是数列各项相加的结果。当我们已知或猜测某个求和公式时，数学归纳法为“该公式对所有正整数 (n) 都成立”提供了严格的证明。这一技巧是 AQA A-Level 数学的核心内容，因为它把清晰的逻辑结构与熟练的代数运算结合起来。

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1. What Is Summation by Induction? | 什么是数学归纳法求和

Summation by induction means proving a formula of the form ( sum_{r=1}^{n} u_r = f(n) ) using the principle of mathematical induction. The left-hand side is a series, and (f(n)) is a function of (n).

数学归纳法求和，是指利用数学归纳法原理证明形如 ( sum_{r=1}^{n} u_r = f(n) ) 的公式。等号左边是一个级数，右边是 (n) 的函数。归纳法不仅证明结果正确，还能帮助我们在验证过程中理解求和公式的递推关系。

• English: The goal is to show that the sum formula holds for all positive integers (n), starting from the first term and extending one term at a time.

• 中文：目标是证明该求和公式对所有正整数 (n) 成立，从第一项出发，一次向后多延一项。

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2. The Principle of Mathematical Induction | 数学归纳法的基本思想

The central idea is often described as a domino effect. Suppose we have a statement (P(n)) about the positive integer (n). If we can show that (P(1)) is true, and that whenever (P(k)) is true then (P(k+1)) must also be true, then (P(n)) is true for all positive integers (n).

数学归纳法的核心思想常被比作“多米诺骨牌效应”。设 (P(n)) 是关于正整数 (n) 的命题。如果我们能证明 (P(1)) 成立，并且“若 (P(k)) 成立，则 (P(k+1)) 必成立”，那么 (P(n)) 对所有正整数 (n) 都成立。

P(1) is true, and P(k) ⇒ P(k+1) for all k ≥ 1, therefore P(n) is true for all n ≥ 1.

In the context of a series, (P(n)) is the equality between the sum (sum_{r=1}^{n} u_r) and a closed-form expression (f(n)).

在级数问题中，(P(n)) 就是“前 (n) 项之和 (sum_{r=1}^{n} u_r) 等于表达式 (f(n))”这个等式。

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3. Standard Notation for Summation | 求和中的常用记号

The capital Greek letter sigma, (Sigma), is used to indicate a sum. The expression (sum_{r=1}^{n} u_r) means (u_1 + u_2 + cdots + u_n).

大写希腊字母 (Sigma) 表示求和。式子 (sum_{r=1}^{n} u_r) 表示 (u_1 + u_2 + cdots + u_n)。

Notation	Meaning	中文含义
(sum_{r=1}^{n} r)	Sum of the first (n) positive integers	前 (n) 个正整数的和
(sum_{r=1}^{n} r^{2})	Sum of the first (n) square numbers	前 (n) 个平方数的和
(sum_{r=1}^{n} ar^{r-1})	Finite geometric series	有限等比级数

When carrying out an induction proof, the key transition is:

在进行归纳证明时，最关键的一步转换是：

(sum_{r=1}^{k+1} u_r = sum_{r=1}^{k} u_r + u_{k+1})

This shows that the new sum is the old sum plus exactly the next term.

这说明新的和等于原来的和再加上紧接着的下一项。

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4. The Five-Step Induction Method | 归纳法五步法

The method can be broken into five clear steps. Following this order prevents mistakes and shows the examiner that you understand the structure.

整个方法可以拆成五个清晰的步骤。按照这个顺序书写可以避免错误，也能让阅卷老师看出你理解归纳法的结构。

• Step 1: State the proposition — Write down the statement (P(n)) clearly for a general positive integer (n).

• 第 1 步：写出命题 — 清楚写出对一般正整数 (n) 成立的命题 (P(n))。

• Step 2: Verify the base case — Show that (P(1)) is true by substituting (n = 1) into both sides.

• 第 2 步：验证基础情况 — 将 (n = 1) 分别代入等式两边，验证 (P(1)) 成立。

• Step 3: Make the inductive assumption — Assume that (P(k)) is true for some positive integer (k).

• 第 3 步：作归纳假设 — 假设对某个正整数 (k)，命题 (P(k)) 成立。

• Step 4: Prove the inductive step — Use (P(k)) to prove that (P(k+1)) is true.

• 第 4 步：证明归纳递推 — 利用 (P(k)) 证明 (P(k+1)) 也成立。

• Step 5: Write the conclusion — State that since the base case is true and the implication holds, the result is true for all positive integers by induction.

• 第 5 步：下结论 — 说明基础情况成立，递推关系成立，因此由数学归纳法可知命题对所有正整数都成立。

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5. Worked Example 1: Sum of the First n Natural Numbers | 例 1：前 n 个自然数之和

We want to prove by induction that:

我们要用归纳法证明：

(sum_{r=1}^{n} r = frac{n(n+1)}{2})

Base case: When (n = 1), the left-hand side is (1). The right-hand side is (1(2)/2 = 1). Therefore (P(1)) is true.

基础情况：当 (n = 1) 时，左边为 (1)，右边为 (1(2)/2 = 1)，所以 (P(1)) 成立。

Inductive assumption: Assume that for some (k geq 1),

归纳假设：假设对某个 (k geq 1)，有

(sum_{r=1}^{k} r = frac{k(k+1)}{2})

Inductive step: Now consider the sum up to (k+1):

归纳递推：现在考虑加到第 (k+1) 项的和：

(sum_{r=1}^{k+1} r = frac{k(k+1)}{2} + (k+1))

Factor out ((k+1)):

提取公因式 ((k+1))：

(sum_{r=1}^{k+1} r = (k+1)left(frac{k}{2} + 1right) = frac{(k+1)(k+2)}{2})

This is exactly the formula with (n = k+1). Hence (P(k) Rightarrow P(k+1)). Since (P(1)) is true and the implication is true, the formula holds for all positive integers (n).

这正是把公式中的 (n) 换成 (k+1) 后的结果，因此 (P(k) Rightarrow P(k+1))。由于 (P(1)) 成立且递推关系成立，所以公式对所有正整数 (n) 都成立。

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6. Worked Example 2: Sum of Squares | 例 2：平方数之和

We prove the standard formula:

我们来证明标准公式：

(sum_{r=1}^{n} r^{2} = frac{n(n+1)(2n+1)}{6})

Base case: For (n = 1), LHS (= 1^{2} = 1). RHS (= 1 times 2 times 3 / 6 = 1). So (P(1)) is true.

基础情况：当 (n = 1) 时，左边 (= 1^{2} = 1)，右边 (= 1 times 2 times 3 / 6 = 1)。所以 (P(1)) 成立。

Inductive assumption: Suppose for some (k geq 1),

归纳假设：假设对某个 (k geq 1)，有

(sum_{r=1}^{k} r^{2} = frac{k(k+1)(2k+1)}{6})

Inductive step: Add the next square term ((k+1)^{2}):

归纳递推：加上下一项平方 ((k+1)^{2})：

(sum_{r=1}^{k+1} r^{2} = frac{k(k+1)(2k+1)}{6} + (k+1)^{2})

Factor carefully:

细致地提取公因式：

(= frac{(k+1)left[k(2k+1) + 6(k+1)right]}{6})

(= frac{(k+1)(2k^{2} + 7k + 6)}{6} = frac{(k+1)(k+2)(2k+3)}{6})

The final expression matches the original formula with (n = k+1). Therefore the formula is true for all positive integers (n).

最终表达式与原公式中 (n = k+1) 时的结果一致。因此该公式对所有正整数 (n) 都成立。

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7. Worked Example 3: Sum of Cubes | 例 3：立方数之和

The famous formula for the sum of the first (n) cubes is:

前 (n) 个立方数之和有一个著名公式：

(sum_{r=1}^{n} r^{3} = left[frac{n(n+1)}{2}right]^{2})

Base case: For (n = 1), LHS (= 1^{3} = 1). RHS (= left[frac{1 times 2}{2}right]^{2} = 1). Base case holds.

基础情况：当 (n = 1) 时，左边 (= 1^{3} = 1)，右边 (= left[frac{1 times 2}{2}right]^{2} = 1)。基础情况成立。

Inductive assumption: Assume that for some (k geq 1),

归纳假设：假设对某个 (k geq 1)，有

(sum_{r=1}^{k} r^{3} = left[frac{k(k+1)}{2}right]^{2})

Inductive step: Add ((k+1)^{3}):

归纳递推：加上 ((k+1)^{3})：

(sum_{r=1}^{k+1} r^{3} = frac{k^{2}(k+1)^{2}}{4} + (k+1)^{3})

Factor out ((k+1)^{2}):

提取公因式 ((k+1)^{2})：

(= (k+1)^{2}left(frac{k^{2}}{4} + k + 1right) = frac{(k+1)^{2}(k+2)^{2}}{4})

This is (left[frac{(k+1)(k+2)}{2}right]^{2}), exactly the formula for (n = k+1). Hence the result follows by induction.

这等于 (left[frac{(k+1)(k+2)}{2}right]^{2})，正是 (n = k+1) 时的公式。因此结论由归纳法成立。

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8. Worked Example 4: Finite Geometric Series | 例 4：有限等比级数

For a geometric series with first term (a) and common ratio (x), the sum of the first (n) terms is:

对于首项为 (a)、公比为 (x) 的等比数列，其前 (n) 项和为：

(sum_{i=0}^{n-1} a x^{i} = frac{a(1-x^{n})}{1-x}, quad x neq 1)

Base case: When (n = 1), the LHS is simply (a). The RHS is (frac{a(1-x)}{1-x} = a). So the base case holds.

基础情况：当 (n = 1) 时，左边就是 (a)，右边为 (frac{a(1-x)}{1-x} = a)。基础情况成立。

Inductive assumption: Assume for some (k geq 1),

归纳假设：假设对某个 (k geq 1)，有

(sum_{i=0}^{k-1} a x^{i} = frac{a(1-x^{k})}{1-x})

Inductive step: Add the next term (a x^{k}):

归纳递推：加上下一项 (a x^{k})：

(sum_{i=0}^{k} a x^{i} = frac{a(1-x^{k})}{1-x} + a x^{k})

Combine into one fraction:

合并成一个分式：

(= frac{a(1-x^{k}) + a x^{k}(1-x)}{1-x} = frac{a(1-x^{k+1})}{1-x})

This matches the formula with (n = k+1). Therefore the geometric summation formula is proved by induction.

这与 (n = k+1) 时的公式一致。因此等比数列求和公式由数学归纳法得证。

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9. Common Mistakes and How to Avoid Them | 常见错误与避免方法

Students often lose marks in induction proofs because of small algebraic slips or omitted logical phrases. The table below shows the most common issues.

学生在归纳法证明中经常因为细小的代数错误或漏写逻辑语句而失分。下表列出了最常见的问题。

Mistake	How to Avoid	中文说明
Forgetting the base case	Always test (n = 1) first; some problems may need (n = 0).	永远先检验 (n = 1)，有些题目可能从 (n = 0) 开始。
Using the result to prove the result	Do not replace the sum with (f(k+1)) before proving it.	不要在证明之前就把和替换成 (f(k+1))。
Algebraic factorisation errors	Write every line clearly and factor common terms slowly.	每一步都写清楚，提取公因式时要耐心。
Missing the logical conclusion	End with: “Therefore, by induction, the statement is true for all positive integers (n).”	结尾必须写：“因此，由数学归纳法可知，命题对所有正整数 (n) 成立。”

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10. When to Use and Not Use Induction | 何时使用与不使用归纳法

Induction is ideal when a summation formula is already given or strongly suggested and you need a proof. It is a verification tool, not a discovery tool.

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